Question

The shortest wavelength of $H$ atom in the Lyman series is $\lambda_{1} .$ The longest wavelength in the Balmer series of $He ^{+}$ is :-

• A. $\frac{5 \lambda_{1}}{9}$
• B. $\frac{27 \lambda_{1}}{5}$
• C. $\frac{9 \lambda_{1}}{5}$
• D. $\frac{36 \lambda_{1}}{5}$

Answer 1

Answer: (C)

As we know $\Delta E =\frac{ hc }{\lambda}$

So $\lambda=\frac{h c}{\Delta E} \,\,\,\,$ for $\lambda$ minimum i.e.

shortest; $\Delta E =$ maximum

for Lyman series $n =1 $ for $\Delta E _{\max }$

Transition must be form $n =\infty$ to $n =1$

So $ \frac{1}{\lambda}= R _{ H } Z ^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)$

$\frac{1}{\lambda}= R _{ H } Z ^{2}(1-0) $

$\frac{1}{\lambda}= R \times(1)^{2} \Rightarrow \lambda_{1}=\frac{1}{ R }$

For longest wavelength $\Delta E =$ minimum for Balmer series $n =3$ to $n =2$ will have $\Delta E$ minimum

for $He ^{+} Z =2$

So $\frac{1}{\lambda_{2}}= R _{ H } \times Z ^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)$

$\frac{1}{\lambda_{2}}= R _{ H } \times 4\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{\lambda_{2}}= R _{ H } \times \frac{5}{9}$

$\lambda_{2}=\lambda_{1} \times \frac{9}{5}$